Answer is none of the above
-70°C
Sink
little
hydrogen bonding
Explanation:
Completing the statements:
Water's boiling point would have been close to -70°C. Ice would sink in water. Water would release little heat to warm land during the winter. Ice is less dense than water because of the hydrogen bonding that forms a hexagonal structure in water.
The unique property of water is as a result of its hydrogen bonding. Water is a polar covalent compound. Like most covalent compound, water would have naturally had a very low boiling point.
The intermolecular forces all hydrogen bonding gives water its unique nature.
Hydrogen bond is formed by an attraction between hydrogen one water water molecule and more electronegative atom on another molecule usually oxygen, nitrogen and fluorine.
They form very strong intermolecular interaction responsible for the behavior of water.
The higher specific heat capacity of water is due to this bond. It absorbs a lot of heat and does not release them on time. This causes water release heat during winter.
Water has a hexagonal shape or structure linking each molecules.
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Answer:
Lithospheric plates are regions of Earth's crust and upper mantle that are fractured into plates that move across a deeper plasticine mantle. Earth's crust is fractured into 13 major and approximately 20 total lithospheric plates.
Explanation:
<em>I hope this helps</em>
The idea here is that you need to figure out how many moles of magnesium chloride,
MgCl
2
, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.
As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.
c
=
n
V
So, how many moles of magnesium chloride must be present in the target solution?
c
=
n
V
⇒
n
=
c
⋅
V
n
=
0.158 M
⋅
250.0
⋅
10
−
3
L
=
0.0395 moles MgCl
2
Now determine what volume of the target solution would contain this many moles of magnesium chloride
c
=
n
V
⇒
V
=
n
c
V
=
0.0395
moles
3.15
moles
L
=
0.01254 L
Rounded to three sig figs and expressed in mililiters, the volume will be
V
=
12.5 mL
So, to prepare your target solution, use a
12.5-mL
sample of the stock solution and add enough water to make the volume of the total solution equal to
250.0 mL
.
This is equivalent to diluting the
12.5-mL
sample of the stock solution by a dilution factor of
20
.